Random variables problem

Question

Let $X, Y, Z$ be non-independent random variables. How is the Sylvester criterion applied using the matrix of covariance?

I have calculated $E[X^2]$ using the formula, but how can I use the Sylvester criterion on the calculated matrix?

Answer 1

I will write $E$ for expectation (instead of $M$ for mean) and will assume that by dispersion $D(X)$, you mean the standard deviation, so that the variance $E[X^2] - E[X]^2 = E[X^2]$ is $1$. The basic idea is that the covariance matrix

$\begin{bmatrix} E[X^2] & E[XY] & E[XZ] \\ E[XY] & E[Y^2] & E[YZ] \\ E[XZ] & E[YZ] & E[Z^2] \\ \end{bmatrix} = \begin{bmatrix} 1 & E[XY] & \frac{1}{2} \\ E[XY] & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \\ \end{bmatrix}$

is positive semi-definite (see e.g. here). By Sylvester's criterion, this means that all three upper left square submatrices have non-negative determinant. In particular, letting $x = E[XY]$, this gives $1 - x^2 \geq 0$ and $x^2 - \frac{1}{2}x - \frac{1}{2} \leq 0$, which yield $-\frac{1}{2} \leq E[XY] \leq 1$. Conversely, any positive semi-definite covariance matrix can be realized by random variables (see e.g. here), so these bounds are sharp.

Answer 2

By Cauchy - Schwarz inequality $|M(XY)| \leq \sqrt {MX^{2}}\sqrt {MY^{2}}=1$. Hence $-1 \leq M(XY) \leq 1$. For an example where $M(XY)=1$ take $X$ with standard normal distribution and $Y=X$; you can always find a $Z$ such that $EXZ=\frac 1 2$ and in this case we also have $EYZ=\frac 1 2$. Hence the largest value of $M(XY)$ is $1$.

Similarly the lowest value of $M(XY)$ is $-1$.